arXiv:1508.06355v1 [math.LO] 26 Aug 2015
NO UNWANTED
UNIVERSALLY BAIRE MORPHISMS
DAN HATHAWAY
Abstract. We show that the usual proof that there are no morphisms, whose constituent maps are Borel, between certain challengeresponse relations generalizes to show that there are no morphisms
whose constituent maps are universally Baire.
1. Morphisms
Let κ be a cardinal. Recall that a set A ⊆ ω ω is κ-universally Baire
(see [2]) iff there exist trees T, S ⊆ <ω ω × <ω δ for some cardinal δ such
that p[T ] = A and in every forcing extension of V by a forcing of size
≤ κ, p[T ] = ω ω − p[S]. A set A ⊆ ω ω is universally Baire iff it is κuniversally Baire for all κ. We make a similar definition for relations on
ω
ω to be universally Baire. We say that a function is universally Baire
iff its graph is. Given a set A which is κ-universally Baire, witnessed
by T and S, and given a forcing of size ≤ κ, we say that the set p[T ]
(as computed in the extension) is what A lifts to.
Let R1 := hω ω, ω ω, R1i and R2 := hω ω, ω ω, R2 i be challenge-response
relations (so R1 , R2 ⊆ ω ω × ω ω). It is natural to ask if there is a
morphism form the first to the second. That is, a pair hφ− , φ+ i of
functions φ− , φ+ : ω ω → ω ω such that
(∀x ∈ ω ω)(∀y ∈ ω ω) φ−(x)R1 y ⇒ xR2 φ+ (y).
In [1] (Theorem 4.15), a situation is given where there can be no such
morphism with either φ− or φ+ Borel. The goal of this document is to
show why in the same situation it is impossible for both φ− and φ+ to
be universally Baire. This leaves open the question of whether one of
φ− or φ+ could be universally Baire..
Consider a challenge-response relation R = hω ω, ω ω, Ri which has
an interpretation in every forcing extension (this happens when the
relation R is universally Baire for example). Given a forcing P, we say
that that P is R-adequate if
1P (∀x ∈ ω ω)(∃y ∈ ω ω ∩ V̌ ) xRy.
1
2
DAN HATHAWAY
Lemma 1.1. Let κ be an infinite cardinal. Let f : ω ω → ω ω be a
function whose graph is κ-universally Baire. Then in every forcing
extension by a poset of size ≤ κ, f lifts to a function defined on all of
ω
ω.
Proof. Let κ be a cardinal and let T, S be trees witnessing that the
graph of f is κ-universally Baire. Let P be a poset of size ≤ κ. Let
G be (V, P)-generic. We want to show that (p[T ])V [G] is the graph
of a total function in V [G]. Given any x ∈ (ω ω)V [G] , we want some
y ∈ (ω ω)V [G] satisfying (x, y) ∈ p[T ].
Towards a contradiction, fix an x ∈ (ω ω)V [G] such that there is no
such corresponding y. From T ⊆ <ω ω × <ω ω × <ω δ we can form Tx ⊆
<ω
ω × <ω δ which is a well-founded tree. Since Tx is well-founded, it has
some rank function σ : Tx → ω1 . Consider the tree W whose nodes are
pairs consisting of an element of <ω ω and a partial attempt to build a
rank function for the corresponding tree from T . We have that (x, σ)
is a path through W . Since W ∈ V and W has a path in V [G], it has
a path in V . Such a path witnesses that f is not total in V , which is
a contradiction.
Using similar reasoning, it can be shown that in V [G] there are not
x, y1 , y2 with y1 6∈ y2 such that (x, y1 ) ∈ p[T ] and (x, y2 ) ∈ p[T ].
We will show the following. The hypothesis arises in practice, for
example the proof that there is no Borel morphism from the splitting
relation to the domination relation (see [1] Theorem 4.15).
Proposition 1.2. Let R1 = hω ω, ω ω, R1i and R2 = hω ω, ω ω, R2i be
challenge-response relations with R1 and R2 universally Baire. Suppose
there is a forcing P which is R1 -adequate but not R2 -adequate. Then
there is no morphism hφ− , φ+ i from R1 to R2 such that both the graph
of φ− and the graph of φ+ are universally Baire.
Proof. Consider a pair of functions hφ− , φ+ i that are universally Baire.
Let Φ be the statement that there exist x1 , x2 , y1 , y2 ∈ ω ω satisfying
the following
1) φ− (x1 ) = x2 ;
2) φ+ (y1 ) = y2 ;
3) x2 R1 y1 ;
4) ¬x1 R2 y2 .
xO 2
R1
y1
¬R2
φ+
φ−
x1
y2 .
NO UNWANTED UNIVERSALLY BAIRE MORPHISMS
3
We claim that Φ is equivalent to a statement which asserts the existence
of a path through a tree in the ground model. The conditions 1)-4)
are all of this form. For example, if R2 = p[T ] = ω ω − p[S], then 4) is
equivalent to saying that (x1 , y2 ) is a path through S.
Note that by the definition of a morphism, if Φ holds, then hφ− , φ+ i
is not a morphism from R1 to R2 . Now since Φ is equivalent to a
statement which asserts the existence of a path through a tree in the
ground model, it is absolute between V and forcing extensions. In
particular, if we show that Φ holds after forcing with P, then we are
done.
Force with P to get V [G]. Let x1 ∈ (ω ω)V [G] witness that P is not
R2 -adequate. That is, there is no y ∈ ω ω ∩ V satisfying x1 R2 y. By
the lemma above, we may speak of φ− (x1 ). Since P is R1 -adequate, let
y1 ∈ ω ω ∩ V satisfy φ− (x1 )R1 y1 . By what we said about x1 , we have
¬x1 R2 φ+ (y1 ). Hence, Φ is satisfied. This completes the proof.
Note that this proposition says that φ− and φ+ cannot both be universally Baire, whereas Theorem 4.15 of [1] says that neither φ− nor
φ+ can be Borel.
2. Weak Morphisms
There is a variant of the notion of morphism which is more general
when one does not assume the Axiom of Choice. The idea is to replace
functions with multiple valued functions.
Definition 2.1. Given challenge-response relations A = hω ω, ω ω, Ai
and B = hω ω, ω ω, Bi, a weak morphism from A to B is a pair hF− , F+ i
of relations F− , F+ ⊆ ω ω × ω ω such that the following are satisfied:
1) (∀b− ∈ ω ω)(∃a− ∈ ω ω) b− F− a− ;
2) (∀a+ ∈ ω ω)(∃b+ ∈ ω ω) a+ F+ b+ ;
3) for all a− , a+ , b− , b+ ∈ ω ω, if b− F− a− , a+ F+ b+ , and a− Aa+ , then
b− Bb+ .
This definition can be remembered by the following picture.
a−
A
F−
b−
a+
F+
B
b+ .
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DAN HATHAWAY
If there is a morphism from A to B, then there is a weak morphism
from A to B. Assuming the Axiom of Choice, the other direction holds
as well. The proof of Proposition 1.2 can be easily modified to prove
the corresponding result for weak morphisms.
References
[1] Andreas Blass. Combinatorial Cardinal Characteristics of the Continuum. In
M. Foreman and A. Kanamori, editors, Handbook of Set Theory Volume 1.
Springer, New York, NY, 2010.
[2] Qi Feng, Menachem Magidor, and W. Hugh Woodin. Universally Baire sets of
reals. In H. Judah, W. Just, and H. Woodin, editors, Set Theory of the Continuum, volume 26 of Mathematical Sciences Research Institute Publications,
pages 203-242, Heidelberg, 1992. Stringer–Verlag.
Mathematics Department, University of Michigan, Ann Arbor, MI
48109–1043, U.S.A.
E-mail address: danhath@umich.edu